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 August 28th, 2015, 03:53 PM #11 Senior Member   Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 391 Thanks: 70 V8 Archie, I don't understand how you arrived at your concluding implication in your first response. Please spell it out for me.
 August 28th, 2015, 04:00 PM #12 Senior Member   Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 391 Thanks: 70 V8 Archie, what you did in your second response is NOT what I did in my original post. I subtracted a sequence from an identical sequence, namely 1/2 + 1/4 + 1/8 ... from 1/2 + 1/4 + 1/8 ...
 August 28th, 2015, 04:10 PM #13 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,664 Thanks: 2644 Math Focus: Mainly analysis and algebra No, you subtracted the first term of one series from the second term of another, the second from the third, the third from the fourth, etc.. Anyway, the first example I gave did subtract equal terms from each other and also returned the wrong answer.
August 28th, 2015, 04:13 PM   #14
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Quote:
 Originally Posted by Timios V8 Archie, I don't understand how you arrived at your concluding implication in your first response. Please spell it out for me.
Which part is that?

August 28th, 2015, 04:28 PM   #15
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 Originally Posted by Timios I don't know what you are doing here. But if you do the analagous thing that I did, using your S and 2S, you would calculate 2S-S as follows: 2S= 2 + 4 + 6 + 8 + ... S= 1 + 2 + 3 + 4 + ... Subtract both sides and you have: S = 1 + 2 + 3 + 4 + ...
The point (which I think is pretty well hammered down by now) is that algebraic manipulations of non-converging sums don't behave well. Many of them give different answers if the series are re-ordered. In this case:
S = 1 + 2 + 3 + 4 + ...

2S = 2(1 + 2 + 3 + 4 +...) = 2 + 4 + 6 + 8 +...

2S - S = (2 + 4 + 6 + 8 + ...) - (1 + 2 + 3 + 4 +...)

2S - S = -1 + (2 - 2) - 3 + (4 - 4) - 5 + (6 - 6) - 7 +... = -1 - 3 - 5 - 7 - ...

Or you could do it v8archie's way and get S = 1/2.

Always make sure your sums converge, and even then be wary of changing the order of your series. These can be very tricky to work with.

-Dan

August 29th, 2015, 12:32 PM   #16
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Quote:
 Originally Posted by Timios We will prove that: $\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}... = 1$ (exactly) Proof: (1) Let n = $\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}...$ Multiply each side of (1) by 2 (2) 2n = $\displaystyle 1 + \frac{1}{2}+\frac{1}{4}+\frac{1}{8}...$ Subtract each side of line (1) from each side of line (2) (3) n = 1
First I multiplied each side of (1) by 2. Is there any problem with the resulting equation (2)?

Secondly, I subtracted n from 2n to get n as the left side of (3).
I still see no problem.

Finally I subtracted $\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}...$ from the $\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}...$ on the right side of (2). This seems to be the problem for some of you.

As far as I can see, in all your counter examples with non-converging series, none of you has subtracted an identical series as has been done above. For that reason, I do not see them as true counter examples.

August 29th, 2015, 01:46 PM   #17
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Quote:
 Originally Posted by Timios First I multiplied each side of (1) by 2. Is there any problem with the resulting equation (2)? Secondly, I subtracted n from 2n to get n as the left side of (3). I still see no problem. Finally I subtracted $\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}...$ from the $\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}...$ on the right side of (2). This seems to be the problem for some of you. As far as I can see, in all your counter examples with non-converging series, none of you has subtracted an identical series as has been done above. For that reason, I do not see them as true counter examples.
It only works for you because the sum of your series converges. The other examples show that for series that do not converge the algebra does not work. We are not saying you are wrong...we are saying that you have to check for convergence before you do any of this.

-Dan

August 29th, 2015, 03:05 PM   #18
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$\displaystyle n = \dfrac{3}{2}+\dfrac{9}{4}+\dfrac{27}{8}+ \ldots$

Quote:
 Originally Posted by Timios First I multiplied each side of (1) by 2/3. Is there any problem with the resulting equation (2)? $\displaystyle n = \dfrac{3}{2}+\dfrac{9}{4}+\dfrac{27}{8}+ \ldots$ $\displaystyle \dfrac{2n}{3} = 1 + \dfrac{3}{2}+\dfrac{9}{4}+\dfrac{27}{8}+ \ldots$ Secondly, I subtracted $\displaystyle n$ from $\displaystyle \dfrac{2n}{3}$ to get $\displaystyle \dfrac{-n}{3}$ as the left side of (3). I still see no problem. Finally I subtracted $\displaystyle \dfrac{3}{2}+\dfrac{9}{4}+\dfrac{27}{8}+ \ldots$ from the $\displaystyle \dfrac{3}{2}+\dfrac{9}{4}+\dfrac{27}{8}+ \ldots$ on the right side of (2). This seems to be the problem for some of you.
Following your steps yields $\displaystyle \dfrac{-n}{3}=1$ or $\displaystyle n=-3$.

Put yourself in for a Fields Medal. You've shown that a bunch of positive numbers adds to $\displaystyle -3$.

August 29th, 2015, 03:18 PM   #19
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Quote:
 Originally Posted by Timios First I multiplied each side of (1) by 2. Is there any problem with the resulting equation (2)? Secondly, I subtracted n from 2n to get n as the left side of (3). I still see no problem. Finally I subtracted $\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}...$ from the $\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}...$ on the right side of (2). This seems to be the problem for some of you. As far as I can see, in all your counter examples with non-converging series, none of you has subtracted an identical series as has been done above. For that reason, I do not see them as true counter examples.
My first example:
\begin{aligned} && S &= \phantom{-1+} 1 - 1 + 1-1+\cdots \\ &\text{multiply by -1} & -S &= -1+ 1 - 1 + 1-1+\cdots \\ &\text{subtract} & 2S &= \phantom{-} 1 \end{aligned}

 August 29th, 2015, 04:16 PM #20 Senior Member   Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 391 Thanks: 70 Thank you, mrtwhs. I am convinced that you have provided a true counter-example. Your sarcasm is also cute. Last edited by Timios; August 29th, 2015 at 04:34 PM.

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